JEE Main 2014 — Physics Electromagnetism

2014mcqhard

The space between the plates of a parallel plate capacitor is filled with a 'dielectric' whose 'dielectric constant' varies with distance as per the relation: $\mathrm{K}(\mathrm{x})=\mathrm{K}_{\mathrm{o}}+\lambda \mathrm{x}(\lambda=\mathrm{a} \text { constant })$ The capacitance $\mathrm{C}$ , of the capacitor, would be related to its vacuum capacitance $\mathrm{C}_{\mathrm{o}}$ for the relation :

Official previous-year question

Held on 12 Apr 2014 · Verified 6 Jul 2026.

Options

A
$\left.\mathrm{C}=\frac{\lambda \mathrm{d}}{\ln \left(1+\mathrm{K}_{\mathrm{o}} \lambda \mathrm{d}\right.} \mathrm{C}_{\mathrm{o}}\right)$
B
$\left.\mathrm{C}=\frac{\lambda}{\mathrm{d} \cdot \ln \left(1+\mathrm{K}_{\mathrm{o}} \lambda \mathrm{d}\right.} \mathrm{C}_{\mathrm{o}}\right)$
C
$\left.\mathrm{C}=\frac{\lambda \mathrm{d}}{\ln \left(1+\lambda \mathrm{d} / \mathrm{K}_{\mathrm{o}}\right.} \mathrm{C}_{\mathrm{o}}\right)$
D
$\left.\mathrm{C}=\frac{\lambda}{\mathrm{d} . \ln \left(1+\mathrm{K}_{\mathrm{o}} / \lambda \mathrm{d}\right.} \mathrm{C}_{\mathrm{o}}\right)$

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Solution

The value of dielectric constant is given as, $\begin{aligned} &\mathrm{K}=\mathrm{K}_0+\lambda \mathrm{x} \\ &\text { And, } \mathrm{V}=\int_0^{\mathrm{d}} \mathrm{Edr} \\ &\mathrm{V}=\int_0^{\mathrm{d}} \frac{\sigma}{\mathrm{K}} \mathrm{dx} \end{aligned}$ $\begin{aligned} &=\sigma \int_0^{\mathrm{d}} \frac{1}{\left(\mathrm{~K}_0+\lambda \mathrm{x}\right.} \mathrm{dx} \\ &=\frac{\sigma}{\lambda}\left[\ln \left(\mathrm{K}_0+\lambda \mathrm{d}-\ln \mathrm{K}_0\right]\right. \\ &=\frac{\sigma}{\lambda} \ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_0}\right) \end{aligned}$ Now it is given that capacitance of vacuum $=\mathrm{C}_0$ . Thus, $\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}$ $\begin{aligned} &=\frac{\sigma . \mathrm{s}}{\mathrm{v}}(\text { Let surface area of plates }=\mathrm{s}) \\ &=\frac{\sigma . \mathrm{s}}{\frac{\sigma}{\lambda} \ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_0}\right)} \end{aligned}$ $=\mathrm{s} \lambda \cdot \frac{\mathrm{d}}{\mathrm{d}} \frac{1}{\ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_0}\right)}\left(\because\right.$ in vacuum $\left.\varepsilon_0=1\right)$ $\mathrm{c}=\frac{\lambda \mathrm{d}}{\ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_0}\right)} \cdot \mathrm{C}_0 \quad\left(\right.$ here, $\left.\mathrm{C}_0=\frac{\mathrm{s}}{\mathrm{d}}\right)$

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Options

$\left.\mathrm{C}=\frac{\lambda \mathrm{d}}{\ln \left(1+\mathrm{K}_{\mathrm{o}} \lambda \mathrm{d}\right.} \mathrm{C}_{\mathrm{o}}\right)$

$\left.\mathrm{C}=\frac{\lambda}{\mathrm{d} \cdot \ln \left(1+\mathrm{K}_{\mathrm{o}} \lambda \mathrm{d}\right.} \mathrm{C}_{\mathrm{o}}\right)$

$\left.\mathrm{C}=\frac{\lambda \mathrm{d}}{\ln \left(1+\lambda \mathrm{d} / \mathrm{K}_{\mathrm{o}}\right.} \mathrm{C}_{\mathrm{o}}\right)$

$\left.\mathrm{C}=\frac{\lambda}{\mathrm{d} . \ln \left(1+\mathrm{K}_{\mathrm{o}} / \lambda \mathrm{d}\right.} \mathrm{C}_{\mathrm{o}}\right)$

Solution

JEE Main 2014 — Physics Electromagnetism

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JEE Main 2014 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism